Oblicz stężenie molowe roztworu kwasu azotowego III o pH = 3 i stopniu dysocjacji elektrolitycznej 
= 15% . PILNE DAM NAJ !!!!!!!!
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pH = -log [H(+)]
[H(+)] = 10
[H(+)] = 10^(-3)
HNO2 ⇔ H(+) + NO2(-)
α =15%
α*[HNO2]=[H(+)]
[HNO2]=10^(-3) / 15% = 6,67*10^(-2) mol/dm3